$ \lim_{x\rightarrow a} f(x)= \lim_{x\rightarrow a} [f(x)]$ then at $x=a$ is there a maxima or minima? 
$$ \lim_{x\rightarrow a} f(x)= \lim_{x\rightarrow a} [f(x)]$$ Where [.] denotes the greatest integer function (floor) function. $f(x)$ is non-constant in the neighborhood of 'a' and is continuous function.

I was able to solve one part which concluded that $\lim_{x\rightarrow a}[f(x)]$ exists only when $f(x)$ either increases or decreases at both sides of 'a'. Also since it is a floor function therefore $ \lim_{x\rightarrow a} f(x)= \lim_{x\rightarrow a} [f(x)]$ has to be an integer.
But how can we check whether it is a point of maxima or minima?
 A: Your conclusion (as stated) is wrong: $\lim_{x\to a}[f(x)]$ may exist (though differ from $\lim_{x\to a}f(x)$) also under other circumstances. For example, if $f(x)=\sin x$, $\lim_{x\to a}[f(x)]$ exists except when $a$ is a multiple of $\pi$.
Nevertheless, let $m=\lim_{x\to a}[f(x)]$. Then for $\epsilon=\frac12$ we find $\delta>0$ such that $m-\frac12<[f(x)]<m+\frac12$ for all $x$ with $0<|x-a|<\delta$.
As there can only be one integer between $m-\frac12$ and $m+\frac12$, we conclude that $[f(x)]$ is constant for $0<|x-a|<\delta$. It follows that 
$[f(x)]=m$ for $0<|x-a|<\delta$.
Note that this means $m\in\Bbb Z$ and then $$\tag1m\le f(x)<m+1$$ for these $x$.
As we also know that $\lim_{x\to a}f(x)=m$ and that $f$ is continuous, we conclude $f(a)=m$. Then $(1)$ tells us that $a$ is a local minimum.
is the limit of an integer-valued function, hence indeed integer $m$. As also $\lim_{x\to a}f(x)=m$, we conclude from the continuity of $f$ that $f(a)=m$. Furthermore, we have $[f(x)]=m$ for $x$ in some neighbourhood of $a$
A: This equality holds if $f$ is continuous at $x=a$ and $f(a)\in \mathbb Z$.
$f(a)$ can be both maximum and minimum.
